Question

A proton with charge 1.602 x 10^-19 C moves at a speed of 300 m/s in a magnetic field at an angle of 65 degrees. If the strength of the magnetic field is 19 T, what would be the magnitude of the force the charge experiences? (1 point) 8.28 x 10^-16 N 13.78 x 10^-15 N 5.09 x 10^-14 N 7.75 x 10^-17 N

Answer 1

Answer:

$F_m= 8.28 \times 10^{-16} N$

Explanation:

The magnitude of the force is $F_m=evB\sin\theta\\\implies F_m=1.602 \times 10^{-19}\times 300\times 19\times\sin 65^0= 8.28 \times 10^{-16} N$